1.

From differential equation of linear S.H.M obtain an expression for acceleration. Velocity and displacement of a particle performing S.H.M. A sononmeter wire 1 meter long weighing 2 g is in resonsnace with a tunning fork of frequency 300 Hz. Find tension in the sonometer wire.

Answer» The differential equation of linear S.H.M. is
`(d^(2)x)/(dt^(2))+(k)/(m)x=0`
When `m to` mass of the particle performing S.H.M. `(d^(2)x)/(dt^(2))to` acceleration of particles,
`k to` force constant.
`:. (d^(2)x)/(dt^(2))+(k)/(m)x=0`
`:. "Let", (k)/(m)= omega^(2)`
`:. (d^(2)x)/(d^(2))+omega^(2)=0`
Acceleration,
`a=(d^(2)x)/(dt^(2))`
`=-omega^(2)x " "...(i)`
This is the expression of acceleration,
Where, negative sign indicate that acceleration and the dissplacement are in oppsite direction.
Now,
`v=(dv)/(dt)`
`:. a=(d^(2)x)/(dt^(2))`
`=(dv)/(dt)=(dv)/(dx)*(dx)/(dt)=(dv)/(dx)*y`
`=v*(dv)/(dx)`
`:. v*(dv)/(dx)=-omega^(2)x`
Intergrating both sides
` int v dv=- omega^(2) int x dx`
`:. (v^(2))/(2)=-(omega^(2)x^(2))/(2)+C" "...(ii)`
where, C is integration constant.
At extreme position
`x=+-A` where, A is Amplitude.
`0=-(omega^(2)A^(2))/(2)+C`
`C=(omega^(2)A^(2))/(2)`
On substituting the value of C in equation (i),
`(v^(2))/(2)=-(omega^(2)x^(2))/(2)+(omega^(2)A^(2))/(2)`
`v^(2)=omega^(2)(A^(2)-x^(2))`
`v=+- omega sqrt(A^(2)-x^(2))`
On Substituting the value of C in equation (i), `(v^(2))/(2)=-(omega^(2)x^(2))/(2)+(omega^(2)A^(2))/(2)`
`v^(2)=omega^(2)(A^(2)-x^(2))`
`v=+-omegasqrt((A^(2)-x^(2))`
This is the expression for velocity
Now,`(dx)/(dt)=omegasqrt(A^(2)-x^(2)) ( :. v=(dv)/(dt))`
`(dx)/(sqrt(A^(2)-x^(2)))= omega dt`
Integrating both side
`int (dx)/(sqrt(A^(2)-x^(2)))int omega. dt`
`sin^(-1)((x)/(A))= omega t+alpha`
where, `alpha` is integration constant.
`:. (x)/(A)= sin (omegat+alpha)`
`x=A sin (omega t+alpha)`
This is the expression for displacement, where `alpha` depends upon internal condition.
Numerical
Given `L=1m,=2g=0.002kg,n=300Hz`.
`n=(1)/(2L)sqrt((T)/(m))`
`T=4n^(2)L^(2)m`
`=3(300)^(2)xx(1)^(2)xx0.002`
`T=720N`


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